Question

What is the second derivative of `f(x)=cos(e^-x)`?

Answer 1

The second derivative is: `f''(x) =e^-x((sin(e^-x)-cos(e^-x))`.

Explanation:

To solve this problem we first need to remember the chain rule, which state that the derivative of `y=f(u)` if `u=g(x)` is
`y'=f'(u)*u'` or `y'=(f'(g(x))*(g'(x))`

In this case we have `y=cos(e^-x)` which is `y=cos(u)` where `u=e^-x`

`f'(x)=-sin(e^-x)*(-e^-x)` So:
`f'(x)=e^-x*sin(e^-x)`
For the second derivative, we use the product and chain rules:

`f''(x)=-e^-x*sin(e^-x)+e^-x*cos(e^-x)`
Factoring, we have:
`f''(x) =e^-x((sin(e^-x)-cos(e^-x))`